Electronic – 5th order low pass filter transfer function

filterlow pass

It is a complete mystery to me how one would write down the transfer function of a generic 5th order low pass filter.

I find it inherently confusing because the question doesn't give any hint or indication as to what the constraints are.

All I can come up with is a five-fold coupled RC low pass filter where we assume that they do no load each other.

Since I've seen examples of a 5th order Chebyshev- and a 5th order Butterworth filter, which are seen to differ in their Bode plots, I sense that the question at hand is simply ill posed.

Best Answer

A low pass filter is as follows:

$$ \frac{Vout}{Vin}= \frac{1}{\tau*s+1} $$

where \$\tau\$ is equal to \$ RC\$.

Since they are linear in the frequency space they multiply:

$$ \frac{Vout}{Vin}= \frac{1}{\tau_1*s+1}*\frac{1}{\tau_2*s+1}*\frac{1}{\tau_3*s+1}*\frac{1}{\tau_4*s+1}*\frac{1}{\tau_5*s+1}\ .$$

If your tau's are all the same then it would be: $$ \frac{Vout}{Vin}= \left( \frac{1}{\tau*s+1} \right)^5\ .$$

Realizing these filters are different, as each RC stage will present a load to the stages after that, which is why we use op amps to isolate the impedance from each stage.